467. Unique Substrings in Wraparound String

Consider the string

s

to be the infinite wraparound string of "abcdefghijklmnopqrstuvwxyz", so

s

will look like this: "...zabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcd....".

Now we have another string

p

. Your job is to find out how many unique non-empty substrings of

p

are present in

s

. In particular, your input is the string

p

and you need to output the number of different non-empty substrings of

p

in the string

s

.

Note:

p

consists of only lowercase English letters and the size of p might be over 10000.

Example 1:

Input: "a"
Output: 1

Explanation: Only the substring "a" of string "a" is in the string s.

Example 2:

Input: "cac"
Output: 2
Explanation: There are two substrings "a", "c" of string "cac" in the string s.

Example 3:

Input: "zab"
Output: 6
Explanation: There are six substrings "z", "a", "b", "za", "ab", "zab" of string "zab" in the string s.

这道题其实就是寻找一个字符串p中有多少个非空子串包含在s中。所求的子串具有连续递增的特点，即类似abcd、zabc而不是acd。解题方法为找出以’a-z’每个字符结尾的情况下，最长的子串有多长，然后将其相加就可以。代码如下：

public
int findSubstringInWraproundString(String p) {

        int p_int[] =
new int[p.length()];

        int count[] =
new int[26];

        for(int i=0;i<p.length();i++){

            p_int[i] = p.charAt(i) - 'a';

        }

        int res = 0;

        int maxLen =
0;

        for( int i=0;i<p.length();i++ ){

            if( i>0 && (p_int[i-1] +
1) % 26 == p_int[i]){

                maxLen ++ ;

            } else{

                maxLen = 1;

            }

            count[p_int[i]] = Math.max(count[p_int[i]],maxLen);

        }

        for( int i=0;i<26;i++)

            res += count[i];

        return res;

    }